2 Answers
2

The general approach described by Todd Wilcox is the best way to go about things: It is important to become thoroughly familiar with the function $e^{rt}$.

When doubling time is given, there is a quick formula that works. Let $d$ be the doubling time. Then if $P(t)$ is the population at time $t$, we have
$$P(t)=P(0)\,2^{t/d}.$$
So in our case,
$$P(t)=(100)2^{t/3}.\tag{$1$}$$
The general approach, which you should carry out, will give you after some manipulation
$$P(t)=(100)e^{t \log 2/3},$$
where by $\log$ we mean logarithm to the base $e$ ($\ln$ on your calculator).
This is the same as the answer $(1)$, since
$$e^{x\log 2}=\left(e^{\log 2}\right)^x=2^x,$$
since $e^{\log 2}=2$.

Remark: As long as one is willing to believe that the answer has shape $P(t)=A\,2^{kt}$, the formula is easy to prove. For put $t=0$. Then $P(0)=A\,2^{(k)(0)}=A$, so $A=P(0)$. Also, since doubling time is $d$, we have $P(d)=2P(0)$. But $P(t)=P(0)\,2^{kt}$. Put $t=d$. We get $2P(0)=P(0)\,2^{kd}$, and therefore $kd=1$, meaning that $k=1/d$. This yields the formula $P(t)=P(0)\,2^{t/d}$ mentioned above.

All exponential growth or decay follow the same basic mathematical relationship:$$y(t)=Pe^{rt},$$where $y$ represents the amount of whatever is growing or decaying at time $t$, $P$ is the "principle" (taken from compounding interest on money) or the starting value, and $r$ is the rate of growth or decay.

Often, as in your question, we have to find $r$ using other known quantites. In your case, we know that if we start with $P$ bacteria, we should have $y(3)=2P$, or double the initial amount after three hours. So we can set up and then solve for $r$:$$2P=Pe^{3r}.$$

Once you have $r$ and the initial amount (100 for the second half of your problem), you can find the final amount at any time $t$. Does that help?